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Let \(A\) be an invertible matrix. If \(\mathbf{V}\) is an eigenvector of \(A\), show it is also an eigenvector of both \(A^{2}\) and \(A^{-2}\). What are the corresponding eigenvalues?
in Mathematics by Platinum (164,236 points) | 280 views

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If $\mathbf{V}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A\mathbf{V} = \lambda \mathbf{V}$.

  1. Eigenvector of $A^2$: We have $$A^2\mathbf{V} = A(A\mathbf{V}) = A(\lambda \mathbf{V}) = \lambda (A\mathbf{V}) = \lambda^2 \mathbf{V}$$ Hence, $\mathbf{V}$ is also an eigenvector of $A^2$ with eigenvalue $\lambda^2$.

  2. Eigenvector of $A^{-2}$: Since $A$ is invertible, $A^{-1}$ exists and $A^{-2} = (A^{-1})^2$. We have $$A^{-2} \mathbf{V} = (A^{-1})^2\mathbf{V} = A^{-1}(A^{-1} \mathbf{V}) = A^{-1}(\frac{1}{\lambda}\mathbf{V}) = \frac{1}{\lambda^2} \mathbf{V}$$. Hence, $\mathbf{V}$ is also an eigenvector of $A^{-2}$ with eigenvalue $\frac{1}{\lambda^2}$.

ago by Diamond (75,023 points)

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MathsGee Android Q&A

MathsGee Android Q&A